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Questions Related to maths

Multiple choice maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

Which of the following matrices is not invertible?

  1. $\displaystyle \left[ \begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix} \right] $
  2. $\displaystyle \left[ \begin{matrix} -1 & -1 \\ -1 & 2 \end{matrix} \right] $
  3. $\displaystyle \left[ \begin{matrix} 2 & 3 \\ 4 & 6 \end{matrix} \right] $
  4. $\displaystyle \left[ \begin{matrix} 2 & -2 \\ 1 & 1 \end{matrix} \right] $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

A square matrix that is not invertible is called singular matrix in which its determinant is 0.
$\displaystyle \left[\begin{matrix} 2 & 3 \ 4 & 6 \end{matrix}  \right]$ is a non invertible matrix.
$\because$ its determinant is    $12-12 =0$
Option C is correct.

Multiple choice maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

Consider two matrix $A = \begin{bmatrix} 1 & 2\ 2 & 1\ 1 & 1 \end{bmatrix}$ and $ B = \begin{bmatrix} 1 & 2 & -4\  2 & 1 & -4 \end{bmatrix}$. Which one of the following is correct ?

  1. B is the right inverse of A

  2. B is the left inverse of A

  3. B is the both sided inverse of A

  4. None of the above
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For two $m \times  n$ and $n \times m$ matrices $A$ and $B$

If $AB =I$, then $B$ is right inverse of $A$
If $BA=I$, then $B$ is left inverse of $A$
Note that if $AB=BA=I$, $B$ is both sided inverse of $A$
If $m=n$, which isn't true here
$AB=\begin{bmatrix} 1 & 2 \ 2 & 1 \ 1 & 1 \end{bmatrix}\begin{bmatrix} 1 & 2 & -4 \ 2 & 1 & -4 \end{bmatrix}=\begin{bmatrix} 5 & 4 & -4 \ 4 & 5 & -4 \ 3 & 3 & 0 \end{bmatrix}\neq I$
$BA=\begin{bmatrix} 1 & 2 & -4 \ 2 & 1 & -4 \end{bmatrix}\begin{bmatrix} 1 & 2 \ 2 & 1 \ 1 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}=I$
$\therefore$ B is left inverse of A

Multiple choice maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

If the sum of infinite G.P. $p, 1, \dfrac{1}{p}, \dfrac{1}{p^2}, ......., $ is $\dfrac{9}{2}$. Then find the value of $p$.

  1. $1$
  2. $\dfrac{3}{2}$
  3. $3$
  4. $\dfrac{5}{2}$
Reveal answer Fill a bubble to check yourself
B,C Correct answer
Explanation

Sum of infinite series of a GP $=\dfrac {a} {1-r}$, where $a$ is the first term and $r$ is the common ratio

 
Here $a=p$ and $r=\dfrac {1}{p}$

$\Rightarrow \dfrac {p} {1-\dfrac {1}{p}}=\dfrac {9}{2}$

$\Rightarrow \dfrac {p^{2}}{p-1}=\dfrac {9}{2}$

$\Rightarrow 2p^{2}-9p+9=0$

$\Rightarrow 2p^{2}-6p-3p+9=0$

$\Rightarrow (2p-3)(p-3)=0$

$\Rightarrow p=3,\dfrac{3}{2}$    (because common ratio $r=1/p$ must be less than 1) 

Multiple choice maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

Sum $1,\sqrt { 3 } ,3......$ to $12$ terms is

  1. $364\left( \sqrt { 3 } +1 \right)$
  2. $364\left( \sqrt { 3 } -1 \right)$
  3. $\dfrac { 364 }{ \left( \sqrt { 3 } -1 \right) } $
  4. $\dfrac { 728 }{ \left( \sqrt { 3 } +1 \right) }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$1\,,\,\,\sqrt 3 \,\,,\,3\,,\,.....12\,\,terms$
  It is a G.P with $a = 1\,,\,r\, = \sqrt 3 $
$1\,,\,\,\sqrt 3 \,\,,\,3\,,\,.....12\,\,terms$
${S _n} = \cfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}$
$ \Rightarrow {S _{12}} = \cfrac{{1\left( {{{\left( {\sqrt 3 } \right)}^{12}} - 1} \right)}}{{\sqrt 3  - 1}}$
$ \Rightarrow {S _{12}} = \cfrac{{{3^6} - 1}}{{\sqrt 3  - 1}} \times \cfrac{{\sqrt 3  + 1}}{{\sqrt 3  + 1}}$
$ \Rightarrow {S _{12}} = \cfrac{{728\left( {\sqrt 3 } \right. + \left. 1 \right)}}{{3 - 1}}$
$\Rightarrow {S _{12}} = \cfrac{{728\left( {\sqrt 3 } \right. + \left. 1 \right)}}{2}$
$ \Rightarrow {S _{12}} = 364\left( {\sqrt 3 } \right. + \left. 1 \right)$
Multiple choice maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

The sum of $2n$ terms of a geometric progression whose first term is $'a'$ and common ratio $'r'$ is equal to the sum of $n$ terms of a geometric progression whose first term is $'b'$ and common '$r^{2}$'. then $b$ is equal to

  1. The sum of the first two terms of the first series.

  2. The sum of the first and last terms of the first series.

  3. The sum of the last two terms of the first series.

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given that

$\begin{array}{l} \dfrac { { a\left( { { r^{ 2n } }-1 } \right)  } }{ { r-1 } } =\dfrac { { b{ { \left( { { r^{ 2 } } } \right)  }^{ n } }-1 } }{ { { r^{ 2 } }-1 } }  \ \Rightarrow \dfrac { { a\left( { { r^{ 2n } }-1 } \right)  } }{ { r-1 } } =\dfrac { { b\left( { { r^{ 2n } }-1 } \right)  } }{ { (r-1)\left( { r+1 } \right)  } }  \ \Rightarrow b=a\left( { r+1 } \right)  \ \Rightarrow b=a+ar \end{array}$
$b$= sum of first two term of the first series.

Multiple choice maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

The value of $x$ that satisfies the relation $x=1-x+x^{2}-x^{3}+x^{4}-x^{5}+.\infty$ if $|x|<1$ 

  1. $\dfrac{-1\pm\sqrt5}{2}$
  2. $\dfrac{-1\pm3i}{2}$
  3. $0$
  4. $none$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Here the first term is $a$ 
Common ratio is given by $-x$
The sum of infinite series is of an GP is given by $\dfrac{a}{1-r}\\x=dfrac{1}{1+x}\\x^2+x=1\\x^2+x-1=0$
Using quadratic formulae $\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$
Here $a=1b=1c=-1$
$\implies x=\dfrac{-1\pm\sqrt{1+4}}{2}\\x=\dfrac{-1\pm\sqrt{5}}{2}$
Multiple choice maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

In a infinite G.P. , the sum of first three terms is 70. If the extreme terms are multiplied by 4 and the middle term is multiplied by 5, the resulting terms form an A.p. then the sum to infinite terms of G.p.   

  1. 120

  2. -40

  3. 160

  4. 80
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We have,

The three numbers be $a,ar\,and\,a{{r}^{2}}$.

Given that,


$ a+ar+a{{r}^{2}}=70 $

$ \Rightarrow a\left( 1+r+{{r}^{2}} \right)=70\,\,......\,\,\left( 1 \right) $


Also given that,

$4a,\,5ar\,and\,4a{{r}^{2}}$ in an A.P.

$\begin{align}

$ \Rightarrow 2\left( 5ar \right)=4a+4a{{r}^{2}} $

$ \Rightarrow 5r=2+2{{r}^{2}} $

$ \Rightarrow 2{{r}^{2}}-5r+2=0 $

$ \Rightarrow 2{{r}^{2}}-\left( 4+1 \right)r+2=0 $

$ \Rightarrow 2{{r}^{2}}-4r-1r+2=0 $

$ \Rightarrow 2r\left( r-2 \right)-1\left( r-2 \right)=0 $

$ \Rightarrow \left( r-2 \right)\left( 2r-1 \right)=0 $

$ \Rightarrow r-2=0,\,\,2r-1=0 $

$ \Rightarrow r=2,\,\,r=\dfrac{1}{2} $

From (1) we get,

$a=10\,\,\,at\,\,\,r=2$

And $a=40\,\,\,at\,\,\,r=\dfrac{1}{2}$

Sum of this series

$ =\dfrac{a}{1-r} $

$ =\dfrac{40}{1-2} $

$ =-40 $

This is the answer.

Multiple choice maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

If a $ >0,  $ then the minimum value of sum of $  \dfrac{1}{a}, 1, a^{2}, a^{3}, \dfrac{1}{a^{4}}  $ is equal to

  1. $2$
  2. $3$
  3. $4$
  4. $5$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Sum of $\dfrac{1}{a},1,a^2,a^3,\dfrac{1}{a^4}$ is $1+a^2+a^3+\dfrac{1}{a}+\dfrac{1}{a^4}$

$AM\ge GM$
$\implies \dfrac{1+a^2+a^3+\frac{1}{a}+\frac{1}{a^4}}{5}\ge \sqrt[5]{(1)(a^2)(a^3)(\frac{1}{a})(\frac{1}{a^4})}$
$1+a^2+a^3+\dfrac{1}{a}+\dfrac{1}{a^4}\ge 5\sqrt[5]{1}$

$1+a^2+a^3+\dfrac{1}{a}+\dfrac{1}{a^4}\ge 5$
The minimum value of $1+a^2+a^3+\dfrac{1}{a}+\dfrac{1}{a^4}$ is $5$